\(\int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx\) [566]

Optimal result

Integrand size = 16, antiderivative size = 82 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\frac {5}{2} \sqrt {x} \sqrt {2-b x}+\frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

5*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(1/2)+5/6*(-b*x+2)^(3/2)*x^(1/2)+1/3*(-b*x+2)^(5/2)*x^(1/2)+5/2*x^(1/2
)*(-b*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {5}{2} \sqrt {x} \sqrt {2-b x} \]

[In]

Int[(2 - b*x)^(5/2)/Sqrt[x],x]

[Out]

(5*Sqrt[x]*Sqrt[2 - b*x])/2 + (5*Sqrt[x]*(2 - b*x)^(3/2))/6 + (Sqrt[x]*(2 - b*x)^(5/2))/3 + (5*ArcSin[(Sqrt[b]
*Sqrt[x])/Sqrt[2]])/Sqrt[b]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5}{3} \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx \\ & = \frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5}{2} \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx \\ & = \frac {5}{2} \sqrt {x} \sqrt {2-b x}+\frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5}{2} \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx \\ & = \frac {5}{2} \sqrt {x} \sqrt {2-b x}+\frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {5}{2} \sqrt {x} \sqrt {2-b x}+\frac {5}{6} \sqrt {x} (2-b x)^{3/2}+\frac {1}{3} \sqrt {x} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\frac {1}{6} \sqrt {x} \sqrt {2-b x} \left (33-13 b x+2 b^2 x^2\right )-\frac {10 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{\sqrt {b}} \]

[In]

Integrate[(2 - b*x)^(5/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(33 - 13*b*x + 2*b^2*x^2))/6 - (10*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])
/Sqrt[b]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95

method	result	size
meijerg	\(\frac {15 \sqrt {-b}\, \left (-\frac {8 \sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {-b}\, \left (\frac {1}{24} b^{2} x^{2}-\frac {13}{48} b x +\frac {11}{16}\right ) \sqrt {-\frac {b x}{2}+1}}{15}-\frac {\sqrt {\pi }\, \sqrt {-b}\, \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3 \sqrt {b}}\right )}{\sqrt {\pi }\, b}\)	\(78\)
default	\(\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(91\)
risch	\(-\frac {\left (2 b^{2} x^{2}-13 b x +33\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{6 \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(104\)

[In]

int((-b*x+2)^(5/2)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

15*(-b)^(1/2)/Pi^(1/2)/b*(-8/15*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(1/2)*(1/24*b^2*x^2-13/48*b*x+11/16)*(-1/2*b*x+1
)^(1/2)-1/3*Pi^(1/2)*(-b)^(1/2)/b^(1/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\left [\frac {{\left (2 \, b^{3} x^{2} - 13 \, b^{2} x + 33 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{6 \, b}, \frac {{\left (2 \, b^{3} x^{2} - 13 \, b^{2} x + 33 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{6 \, b}\right ] \]

[In]

integrate((-b*x+2)^(5/2)/x^(1/2),x, algorithm="fricas")

[Out]

[1/6*((2*b^3*x^2 - 13*b^2*x + 33*b)*sqrt(-b*x + 2)*sqrt(x) - 15*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sq
rt(x) + 1))/b, 1/6*((2*b^3*x^2 - 13*b^2*x + 33*b)*sqrt(-b*x + 2)*sqrt(x) - 30*sqrt(b)*arctan(sqrt(-b*x + 2)/(s
qrt(b)*sqrt(x))))/b]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.24 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.52 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\begin {cases} \frac {i b^{3} x^{\frac {7}{2}}}{3 \sqrt {b x - 2}} - \frac {17 i b^{2} x^{\frac {5}{2}}}{6 \sqrt {b x - 2}} + \frac {59 i b x^{\frac {3}{2}}}{6 \sqrt {b x - 2}} - \frac {11 i \sqrt {x}}{\sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {7}{2}}}{3 \sqrt {- b x + 2}} + \frac {17 b^{2} x^{\frac {5}{2}}}{6 \sqrt {- b x + 2}} - \frac {59 b x^{\frac {3}{2}}}{6 \sqrt {- b x + 2}} + \frac {11 \sqrt {x}}{\sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

integrate((-b*x+2)**(5/2)/x**(1/2),x)

[Out]

Piecewise((I*b**3*x**(7/2)/(3*sqrt(b*x - 2)) - 17*I*b**2*x**(5/2)/(6*sqrt(b*x - 2)) + 59*I*b*x**(3/2)/(6*sqrt(
b*x - 2)) - 11*I*sqrt(x)/sqrt(b*x - 2) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b), Abs(b*x) > 2), (-b**3*x
**(7/2)/(3*sqrt(-b*x + 2)) + 17*b**2*x**(5/2)/(6*sqrt(-b*x + 2)) - 59*b*x**(3/2)/(6*sqrt(-b*x + 2)) + 11*sqrt(
x)/sqrt(-b*x + 2) + 5*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.37 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=-\frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} + \frac {\frac {15 \, \sqrt {-b x + 2} b^{2}}{\sqrt {x}} + \frac {40 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {33 \, {\left (-b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{3} - \frac {3 \, {\left (b x - 2\right )} b^{2}}{x} + \frac {3 \, {\left (b x - 2\right )}^{2} b}{x^{2}} - \frac {{\left (b x - 2\right )}^{3}}{x^{3}}\right )}} \]

[In]

integrate((-b*x+2)^(5/2)/x^(1/2),x, algorithm="maxima")

[Out]

-5*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/sqrt(b) + 1/3*(15*sqrt(-b*x + 2)*b^2/sqrt(x) + 40*(-b*x + 2)^(3/2)
*b/x^(3/2) + 33*(-b*x + 2)^(5/2)/x^(5/2))/(b^3 - 3*(b*x - 2)*b^2/x + 3*(b*x - 2)^2*b/x^2 - (b*x - 2)^3/x^3)

Giac [A] (verification not implemented)

none

Time = 5.75 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b} - \frac {5}{b}\right )} + \frac {15}{b}\right )} + \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} b}{6 \, {\left | b \right |}} \]

[In]

integrate((-b*x+2)^(5/2)/x^(1/2),x, algorithm="giac")

[Out]

1/6*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)*(2*(b*x - 2)/b - 5/b) + 15/b) + 30*log(abs(-sqrt(-b*x +
2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/sqrt(-b))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2-b x)^{5/2}}{\sqrt {x}} \, dx=\int \frac {{\left (2-b\,x\right )}^{5/2}}{\sqrt {x}} \,d x \]

[In]

int((2 - b*x)^(5/2)/x^(1/2),x)

[Out]

int((2 - b*x)^(5/2)/x^(1/2), x)